On Wed, 02 Feb 2005 14:45:35 -0800 (PST), Simo Melenius <[EMAIL PROTECTED]> wrote: > Philippe Fremy <[EMAIL PROTECTED]> writes: > > > You can not reproduce the C++ overload idiom but you can get something > > close with manual type testing. > > > > > To in a > > > function do an if statement with the type() function? > > > > I am not aware of any other method. > > > > def a( arg1 ): > > if type(arg1) == types.IntType: return aWithInt(arg1) > > if type(arg1) == types.ListType: return aWithList(arg1) > > ... > > Or: > > def a_overloader (arg1): > return my_typed_a_func_dict[type (arg1)] (arg1) > > Next I'd put my hands in automating the creation of these wrappers and > the my_typed_a_func_dict map based on my implementation written so > far. Then I'd think of parameterizing on any arbitrary destructuring > of arguments like def foo ((a,b), c) and also on classes to which > instance methods are bound to. At this point, a preprocessor might be > handy to avoid seeing all the internals after which things would > probably start looking sick enough to either switch languages or > thinking of the right problem first and then come up with a pythonic > solution to _that_.
Pah. Use a decorator. Code is available here: Jack Diederich posted a good multimethod snippet here, in response to my own clumsy effort. http://groups-beta.google.com/group/comp.lang.python/browse_thread/thread/5b50153e3fb84862 http://tinyurl.com/4awat type(arg1) == types.IntType def multi_type(t): return multi(lambda x:type(x) == t) @multi_type(types.IntType) def foo(x): return someOperationOnInt(x) @multi_type(types.FloatType) def foo(x): return someOperationOnFloat(x) Regards, Stephen Thorne. "Just because it is possible doesn't make a good idea *wink*", -Jack Diederich, Aug 2004 -- http://mail.python.org/mailman/listinfo/python-list
